Suppose that $f: \mathbb{R} \to \mathbb{R}$ is differentiable at $c$ and that $f(c)=0$. Show that $g(x)=|f(x)|$is differentiable at $c$ iff $f'(c)=0$


Suppose $g'(c)$ exists

Then we have that for $x<c$, $\lim_{x \to c}\frac{g(x)-g(c)}{x-c}$ =$\lim_{x \to c}\frac{|f(x)|}{x-c} \le 0$

and for $x>c$, $\lim_{x \to c}\frac{g(x)-g(c)}{x-c}$ =$\lim_{x \to c}\frac{|f(x)|}{x-c} \ge 0$

Then, by squeeze theorem, we have that $g'(c)=0$

Can anyone please tell me how to use the aforementioned result to get to the point where I can show that $f'(c)=0$?


$f'(c) = 0 \implies \lim_{x \to c}\frac{f(x)}{x-c} =0$ $\implies |f'(c)|=0 = g'(c)$

Can anyone please verify this proof and also help arrive at the result that is emboldened? I'd also appreciate if you could lend some tips/hints to approach such questions and if there are easier ways to proof this.

Thank you.



$$ \left|\frac{|f(x)| -|f(c)|}{x-c} - 0\right|= \frac{|f(x)|}{|x-c|} = \left|\frac{f(x) - f(c)}{x-c}- 0\right|$$

  • $\begingroup$ Thank you! That's using through the $\epsilon - \delta$ definition of a derivative right? Also, could you please comment on whether or not the rest of my proof is correct? $\endgroup$
    – Alea
    May 15 '18 at 9:53
  • 1
    $\begingroup$ You showed correctly that $\lim_{x \to c} \frac{|f(x)|}{x-c} = 0$ and just asserted $f'(c) = 0$. To be complete you would need to show more steps -- using my hint -- or noting that $\frac{|f(x)|}{x-c} \to 0$ if and only if $\frac{|f(x)|}{|x-c|} \to 0$ and $\frac{|f(x)|}{|x-c|} = \left|\frac{f(x)}{x-c}\right| = \left|\frac{f(x)-f(c)}{x-c}\right| $. $\endgroup$
    – RRL
    May 15 '18 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.